The-flow-velocity-of-a-river-increases-linearly-with-the-distance-r-from-its-bank-and-has-its-maximum-value-v-0-in-the-middle-of-the-river-The-velocity-near-the-bank-is-zero-A-boat-which-can-move

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Question Number 18269 by Tinkutara last updated on 17/Jul/17

$$\mathrm{The}\mathrm{flow}\mathrm{velocity}\mathrm{of}\mathrm{a}\mathrm{river}\mathrm{increases}$$$$\mathrm{linearly}\mathrm{with}\mathrm{the}\mathrm{distance}\left(r\right)\mathrm{from}\mathrm{its}$$$$\mathrm{bank}\mathrm{and}\mathrm{has}\mathrm{its}\mathrm{maximum}\mathrm{value}{v}_0\mathrm{in}$$$$\mathrm{the}\mathrm{middle}\mathrm{of}\mathrm{the}\mathrm{river}.\mathrm{The}\mathrm{velocity}$$$$\mathrm{near}\mathrm{the}\mathrm{bank}\mathrm{is}\mathrm{zero}.\mathrm{A}\mathrm{boat}\mathrm{which}\mathrm{can}$$$$\mathrm{move}\mathrm{with}\mathrm{speed}u\mathrm{in}\mathrm{still}\mathrm{water}\mathrm{moves}$$$$\mathrm{in}\mathrm{the}\mathrm{river}\mathrm{in}\mathrm{such}\mathrm{a}\mathrm{way}\mathrm{that}\mathrm{it}\mathrm{is}$$$$\mathrm{always}\mathrm{perpendicular}\mathrm{to}\mathrm{the}\mathrm{flow}\mathrm{of}$$$$\mathrm{current}.\mathrm{Find}$$$$\left(\mathrm{i}\right)\mathrm{The}\mathrm{distance}\mathrm{along}\mathrm{the}\mathrm{bank}\mathrm{through}$$$$\mathrm{which}\mathrm{boat}\mathrm{is}\mathrm{carried}\mathrm{away}\mathrm{by}\mathrm{the}\mathrm{flow}$$$$\mathrm{current},\mathrm{when}\mathrm{the}\mathrm{boat}\mathrm{crosses}\mathrm{the}$$$$\mathrm{river}.$$$$\left(\mathrm{ii}\right)\mathrm{The}\mathrm{equation}\mathrm{of}\mathrm{trajectory}\mathrm{for}\mathrm{the}$$$$\mathrm{coordinate}\mathrm{system}\mathrm{shown}.\mathrm{Assume}$$$$\mathrm{that}\mathrm{the}\mathrm{swimmer}\mathrm{starts}\mathrm{from}\mathrm{origin}.$$

Commented by Tinkutara last updated on 17/Jul/17

Commented by mrW1 last updated on 19/Jul/17

$${\mathrm{u}}_{\mathrm{y}}=\mathrm{u}$$$$\mathrm{y}=\mathrm{ut}$$$${\mathrm{u}}_{\mathrm{x}}=\frac{2\mathrm{y}}{\mathrm{d}}{\mathrm{v}}_0=\frac{{2\mathrm{v}}_0}{\mathrm{d}}\mathrm{y}\mathrm{with}0⩽\mathrm{y}⩽\frac{\mathrm{d}}{2}$$$${\mathrm{u}}_{\mathrm{x}}=\frac{2(\mathrm{d}-\mathrm{y})}{\mathrm{d}}{\mathrm{v}}_0={2\mathrm{v}}_0(1-\frac{\mathrm{y}}{\mathrm{d}})\mathrm{with}\frac{\mathrm{d}}{2}⩽\mathrm{y}⩽\mathrm{d}$$$$$$$${\mathrm{u}}_{\mathrm{x}}=\frac{\mathrm{dx}}{\mathrm{dt}}=\frac{\mathrm{dx}}{\mathrm{dy}}\times \frac{\mathrm{dy}}{\mathrm{dt}}=\mathrm{u}\frac{\mathrm{dx}}{\mathrm{dy}}$$$$\Rightarrow \mathrm{u}\frac{\mathrm{dx}}{\mathrm{dy}}=\frac{{2\mathrm{v}}_0}{\mathrm{d}}\mathrm{y}\mathrm{with}0⩽\mathrm{y}⩽\frac{\mathrm{d}}{2}$$$$\Rightarrow \mathrm{u}\frac{\mathrm{dx}}{\mathrm{dy}}={2\mathrm{v}}_0(1-\frac{\mathrm{y}}{\mathrm{d}})\mathrm{with}\frac{\mathrm{d}}{2}⩽\mathrm{y}⩽\mathrm{d}$$$$\mathrm{x}=\frac{{\mathrm{v}}_0}{\mathrm{ud}}{\mathrm{y}}^2+{\mathrm{C}}_1$$$${\mathrm{C}}_1=0\mathrm{since}\mathrm{x}=0\mathrm{at}\mathrm{y}=0$$$$\Rightarrow \mathrm{x}=\frac{{\mathrm{v}}_0}{\mathrm{du}}{\mathrm{y}}^2\mathrm{with}0⩽\mathrm{y}⩽\frac{\mathrm{d}}{2}$$$$\mathrm{at}\mathrm{y}=\frac{\mathrm{d}}{2}:\mathrm{x}=\frac{{\mathrm{v}}_0\mathrm{d}}{4\mathrm{u}}$$$$$$$$\mathrm{x}=\frac{{2\mathrm{v}}_0}{\mathrm{u}}(\mathrm{y}-\frac{{\mathrm{y}}^2}{2\mathrm{d}})+{\mathrm{C}}_2$$$$\mathrm{at}\mathrm{y}=\frac{\mathrm{d}}{2}:$$$$\frac{{\mathrm{v}}_0\mathrm{d}}{4\mathrm{u}}=\frac{{2\mathrm{v}}_0}{\mathrm{u}}[\frac{\mathrm{d}}{2}-\frac{\mathrm{d}}{8}]+{\mathrm{C}}_2$$$${\mathrm{C}}_2=-\frac{{\mathrm{v}}_0\mathrm{d}}{2\mathrm{u}}$$$$\Rightarrow \mathrm{x}=\frac{{2\mathrm{v}}_0}{\mathrm{u}}(\mathrm{y}-\frac{{\mathrm{y}}^2}{2\mathrm{d}}-\frac{\mathrm{d}}{4})\mathrm{with}\frac{\mathrm{d}}{2}⩽\mathrm{y}⩽\mathrm{d}$$$$\mathrm{at}\mathrm{y}=\mathrm{d}:\mathrm{x}=\frac{{\mathrm{v}}_0\mathrm{d}}{2\mathrm{u}}$$$$$$$$\mathrm{y}=\mathrm{ut}$$$$\mathrm{x}=\frac{{\mathrm{v}}_0\mathrm{u}}{\mathrm{d}}{\mathrm{t}}^2\mathrm{with}0⩽\mathrm{t}⩽\frac{\mathrm{d}}{2\mathrm{u}}$$$$\mathrm{x}={2\mathrm{v}}_0(\mathrm{t}-\frac{{\mathrm{ut}}^2}{2\mathrm{d}}-\frac{\mathrm{d}}{4\mathrm{u}})\mathrm{with}\frac{\mathrm{d}}{2\mathrm{u}}⩽\mathrm{t}⩽\frac{\mathrm{d}}{\mathrm{u}}$$

Commented by mrW1 last updated on 19/Jul/17

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